Finite Element Methods Lectures_Lui.pdf

8/21/2019 Finite Element Methods Lectures_Lui.pdf

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Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction

© 1998 Yijun Liu, University of Cincinnati 1

Chapter 1. I ntroduction

I . Basic Concepts

The finite element method (FEM), or finite element analysis

(FEA), is based on the idea of building a complicated object with

simple blocks, or, dividing a complicated object into small and

manageable pieces. Application of this simple idea can be found

everywhere in everyday life as well as in engineering.

Examples:

• Lego (kids’ play)

• Buildings

• Approximation of the area of a circle:

Area of one triangle: S Ri i=1

2

2sinθ

Area of the circle: S S R N N

R as N N ii

N

= =

→ → ∞=

∑1

2 21

2

2sin

ππ

where N = total number of triangles (elements).

R

θi

“Element” S i


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Why F ini te Element M ethod?

• Design analysis: hand calculations, experiments, and

computer simulations

• FEM/FEA is the most widely applied computer simulationmethod in engineering

• Closely integrated with CAD/CAM applications

• ...

Appli cations of FEM in Engineer ing

• Mechanical/Aerospace/Civil/Automobile Engineering

•Structure analysis (static/dynamic, linear/nonlinear)

• Thermal/fluid flows

• Electromagnetics

• Geomechanics

• Biomechanics

• ...

Examples:

...


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A Brief H istory of the FEM

• 1943 ----- Courant (Variational methods)• 1956 ----- Turner, Clough, Martin and Topp (Stiffness)

• 1960 ----- Clough (“Finite Element”, plane problems)

• 1970s ----- Applications on mainframe computers

• 1980s ----- Microcomputers, pre- and postprocessors

• 1990s ----- Analysis of large structural systems


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FEM in Structural Analysis

Procedures:

• Divide structure into pieces (elements with nodes)

• Describe the behavior of the physical quantities on eachelement

• Connect (assemble) the elements at the nodes to form anapproximate system of equations for the whole structure

• Solve the system of equations involving unknownquantities at the nodes (e.g., displacements)

• Calculate desired quantities (e.g., strains and stresses) atselected elements

Example:


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Computer Implementations

• Preprocessing (build FE model, loads and constraints)

• FEA solver (assemble and solve the system of equations)

• Postprocessing (sort and display the results)

Available Commercial FEM Software Packages

• ANSYS (General purpose, PC and workstations)

• SDRC/I-DEAS (Complete CAD/CAM/CAE package)

• NASTRAN (General purpose FEA on mainframes)

• ABAQUS (Nonlinear and dynamic analyses)

• COSMOS (General purpose FEA)

• ALGOR (PC and workstations)

• PATRAN (Pre/Post Processor)

• HyperMesh (Pre/Post Processor)

• Dyna-3D (Crash/impact analysis)

• ...


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Objectives of This FEM Course

• Understand the fundamental ideas of the FEM

• Know the behavior and usage of each type of elementscovered in this course

• Be able to prepare a suitable FE model for given problems

• Can interpret and evaluate the quality of the results (knowthe physics of the problems)

• Be aware of the limitations of the FEM (don’t misuse theFEM - a numerical tool)


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I I . Review of Matrix Algebra

L inear System of Algebraic Equations a x a x a x b

a x a x a x b

a x a x a x b

n n

n n

n n nn n n

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

+ + + =

+ + + =

+ + + =

...

...

.......

...

(1)

where x1 , x2 , ..., xn are the unknowns.

In matrix form:

Ax b= (2)

where

[ ]

{ } { }

A

x b

= =

= =

= =

a

a a a

a a a

a a a

x

x

x

x

b

b

b

b

ij

n

n

n n nn

i

n

i

n

11 12 1

21 22 2

1 2

1

2

1

2

...

...

... ... ... ...

...

: :

(3)

A is called a n×n (square) matrix, and x and b are (column)

vectors of dimension n.


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Row and Column Vectors

[ ]v w= =

v v v

w

w

w1 2 3

1

2

3

Matrix Addition and Subtraction

For two matrices A and B, both of the same size (m×n), the

addition and subtraction are defined by

C A B

D A B

= + = +

= − = −

with

with

c a b

d a b

ij ij ij

ij ij ij

Scalar M ul tipli cation

[ ]λ λA = a ij

Matrix Multiplication

For two matrices A (of size l×m) and B (of size m×n), the

product of AB is defined by

C AB= = ∑=

with c a bij ik

k

m

kj

1

where i = 1, 2, ..., l ; j = 1, 2, ..., n.

Note that, in general, AB BA≠ , but ( ) ( )AB C A BC=(associative).


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Transpose of a Matri x

If A = [aij], then the transpose of A is

[ ]AT

jia=

Notice that ( )AB B AT T T = .

Symmetr ic Matrix

A square (n×n) matrix A is called symmetric, if

A A=T or a aij ji=

Uni t (I dentity) Matrix

I =

1 0 0

0 1 0

0 0 1

...

...

... ... ... ...

...

Note that AI = A, Ix = x.

Determinant of a Matri x

The determinant of square matrix A is a scalar number denoted by det A or |A|. For 2×2 and 3×3 matrices, their

determinants are given by

deta b

c d ad bc

= −


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and

det

a a a

a a a

a a a

a a a a a a a a a

a a a a a a a a a

11 12 13

21 22 23

31 32 33

11 22 33 12 23 31 21 32 13

13 22 31 12 21 33 23 32 11

= + +

− − −

Singular Matrix

A square matrix A is singular if det A = 0, which indicates

problems in the systems (nonunique solutions, degeneracy, etc.)

Matr ix I nversion

For a square and nonsingular matrix A (detA ≠ 0), itsinverse A-1 is constructed in such a way that

AA A A I

− −

= =

1 1

The cofactor matrix C of matrix A is defined by

C M ij

i j

ij= − +( )1

where M ij is the determinant of the smaller matrix obtained by

eliminating the ith row and jth column of A.

Thus, the inverse of A can be determined by

AAC

− =11

det

T

We can show that ( )AB B A− − −=1 1 1.


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Examples:

(1)a b

c d ad bc

d b

c a

=

−

−

−

−11

( )

Checking,

a b

c d

a b

c d ad bc

d b

c a

a b

c d

= −

−

−

=

−11 1 0

0 1( )

(2)

1 1 0

1 2 1

0 1 2

1

4 2 1

3 2 1

2 2 1

1 1 1

3 2 1

2 2 1

1 1 1

1−

− −

−

=− −

=

−

( )

T

Checking,

1 1 0

1 2 1

0 1 2

3 2 1

2 2 1

1 1 1

1 0 0

0 1 0

0 0 1

−

− −

−

=

If det A = 0 (i.e., A is singular), then A-1 does not exist!

The solution of the linear system of equations (Eq.(1)) can be

expressed as (assuming the coefficient matrix A is nonsingular)

x A b= −1

Thus, the main task in solving a linear system of equations is to

found the inverse of the coefficient matrix.


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Solution Techniques for L inear Systems of Equations

• Gauss elimination methods

• Iterative methods

Positive Defini te Matr ix

A square (n×n) matrix A is said to be positive definite, if for

any nonzero vector x of dimension n,

x AxT >

0 Note that positive definite matrices are nonsingular.

Di ff erentiation and I ntegration of a Matr ix

Let

[ ]A( ) ( )t a t ij=then the differentiation is defined by

d

dt t

da t

dt

ijA( )

( )=

and the integration by

A( ) ( )t dt a t dt ij

=

∫ ∫


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Types of F inite Elements

1-D (L ine) Element

(Spring, truss, beam, pipe, etc.)

2-D (Plane) Element

(Membrane, plate, shell, etc.)

3-D (Sol id) Element

(3-D fields - temperature, displacement, stress, flow velocity)


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I I I . Spr ing Element

“Everything important is simple .”

One Spring Element

Two nodes: i, j

Nodal displacements: ui , u j (in, m, mm)

Nodal forces: f i , f j (lb, Newton)

Spring constant (stiffness): k (lb/in, N/m, N/mm)

Spring force-displacement relationship:

F k = ∆ with ∆ = −u u j i

k F = / ∆ (> 0) is the force needed to produce a unit stretch.

k

i

u juii j

∆

onlinear

Linear

k


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We only consider linear problems in this introductory

course.

Consider the equilibrium of forces for the spring. At node i,

we have

f F k u u ku kui j i i j= − = − − = −( )

and at node j,

f F k u u ku ku j j i i j

= = − = − +( )

In matrix form,

k k

k k

u

u

f

f

i

j

i

j

−

−

=

or,

ku f =

where

k = (element) stiffness matrix

u = (element nodal) displacement vector

f = (element nodal) force vector

Note that k is symmetric. Is k singular or nonsingular? That is,

can we solve the equation? If not, why?


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Spring System

For element 1,

k k

k k

u

u

f

f

1 1

1 1

1

2

1

1

21

−

−

=

element 2,

k k

k k

u

u

f

f

2 2

2 2

2

3

1

2

2

2

−

−

=

where f im

is the (internal) force acting on local node i of elementm (i = 1, 2).

Assemble the stiffness matrix for the whole system:

Consider the equilibrium of forces at node 1,

F f 1 1

1=

at node 2,

F f f 2 2

1

1

2= +

and node 3,

F f 3 2

2=

k 1

u1, 1

k 2

u2, F 2 u3, 3

1 2 3


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That is,

F k u k u

F k u k k u k u

F k u k u

1 1 1 1 2

2 1 1 1 2 2 2 3

3 2 2 2 3

= −

= − + + −

= − +

( )

In matrix form,

k k

k k k k

k k

u

u

u

F

F

F

1 1

1 1 2 2

2 2

1

2

3

1

2

3

0

0

−

− + −

−

=

or

KU F=

K is the stiffness matrix (structure matrix) for the spring system.

An alternative way of assembling the whole stiffness matrix:

“Enlarging” the stiffness matrices for elements 1 and 2, we

have

k k

k k

u

u

u

f

f

1 1

1 1

1

2

3

1

1

2

1

0

0

0 0 0 0

−

−

=

0 0 0

0

0

0

2 2

2 2

1

2

3

1

2

2

2

k k

k k

u

u

u

f

f

−

−

=


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Adding the two matrix equations ( superposition), we have

k k

k k k k k k

u

uu

f

f f f

1 1

1 1 2 2

2 2

1

2

3

1

1

2

1

1

2

2

2

0

0

−

− + −

−

= +

This is the same equation we derived by using the force

equilibrium concept.

Boundary and load conditions:

Assuming, u F F P 1 2 30= = =and

we have

k k

k k k k

k k

u

u

F

P

P

1 1

1 1 2 2

2 2

2

3

10

0

0−

− + −

−

=

which reduces to

k k k

k k

u

u

P

P

1 2 2

2 2

2

3

+ −

−

=

and

F k u1 1 2= −

Unknowns are

U =

u

u

2

3

and the reaction force F 1 (if desired).


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Solving the equations, we obtain the displacements

u

u

P k

P k P k

2

3

1

1 2

2

2

=+

/

/ /

and the reaction force

F P 1 2= −

Checking the Resul ts

• Deformed shape of the structure

• Balance of the external forces

• Order of magnitudes of the numbers

Notes About the Spring Elements

• Suitable for stiffness analysis

• Not suitable for stress analysis of the spring itself

• Can have spring elements with stiffness in the lateraldirection, spring elements for torsion, etc.


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Example 1.1

Given: For the spring system shown above,

k k k

P u

1 2 3

4 0

= = =

= = =

100 N / mm, 200 N / mm, 100 N / mm

500 N, u1

Find : (a) the global stiffness matrix

(b) displacements of nodes 2 and 3

(c) the reaction forces at nodes 1 and 4

(d) the force in the spring 2

Solution :

(a) The element stiffness matrices are

k 1

100 100

100 100=

−

−

(N/mm) (1)

k 2 200 200

200 200= −−

(N/mm) (2)

k 3

100 100

100 100=

−

−

(N/mm) (3)

k 1 k 2

1 2 3

k 3

4


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Applying the superposition concept, we obtain the global stiffness

matrix for the spring system as

u u u u1 2 3 4

100 100 0 0

100 100 200 200 0

0 200 200 100 100

0 0 100 100

K =

−

− + −

− + −

−

or

K =

−

− −

− −

−

100 100 0 0

100 300 200 0

0 200 300 100

0 0 100 100

which is symmetric and banded .

Equilibrium (FE) equation for the whole system is

100 100 0 0

100 300 200 0

0 200 300 100

0 0 100 100

0

1

2

3

4

1

4

−

− −

− −

−

=

u

u

u

u

F

P

F

(4)

(b) Applying the BC (u u1 4

0= = ) in Eq(4), or deleting the 1st and

4th rows and columns, we have


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300 200

200 300

02

3

−

−

=

u

u P (5)

Solving Eq.(5), we obtainu

u

P

P

2

3

250

3 500

2

3

=

=

/

/( )mm (6)

(c) From the 1st and 4th equations in (4), we get the reaction forces

F u1 2100 200= − = − (N)

F u4 3

100 300= − = − ( ) N

(d) The FE equation for spring (element) 2 is

200 200

200 200

−

−

=

u

u

f

f

i

j

i

j

Here i = 2, j = 3 for element 2. Thus we can calculate the spring

force as

[ ]

[ ]

F f f u

u j i

= = − = −

= −

=

200 200

200 2002

3

200

2

3

(N)

Check the results!


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Example 1.2

Problem : For the spring system with arbitrarily numbered nodes

and elements, as shown above, find the global stiffness

matrix.

Solution :

First we construct the following

which specifies the global node numbers corresponding to the

local node numbers for each element.

Then we can write the element stiffness matrices as follows

k 1

k 242

3

k 3

5

2

1k 4

1

1

2 3

4

Element Connectivity Table

Element Node i (1) Node j (2)

1 4 2

2 2 3

3 3 5

4 2 1


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u u

k k

k k

4 2

1

1 1

1 1

k = −

−

u u

k k

k k

2 3

2

2 2

2 2

k = −

−

u u

k k

k k

3 5

3

3 3

3 3

k = −

−

u u

k k

k k

2 1

4

4 4

4 4

k = −

−

Finally, applying the superposition method, we obtain the globalstiffness matrix as follows

u u u u u

k k

k k k k k k

k k k k

k k

k k

1 2 3 4 5

4 4

4 1 2 4 2 1

2 2 3 3

1 1

3 3

0 0 0

0

0 0

0 0 0

0 0 0

K =

−

− + + − −

− + −

−−

The matrix is symmetric, banded , but singular .


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Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements


Chapter 2. Bar and Beam Elements.

L inear Static Analysis

I . Linear Static Analysis

Most structural analysis problems can be treated as linear

static problems, based on the following assumptions

1. Small deformations (loading pattern is not changed dueto the deformed shape)

2. Elastic materials (no plasticity or failures)

3. Static loads (the load is applied to the structure in a slow

or steady fashion)

Linear analysis can provide most of the information about

the behavior of a structure, and can be a good approximation for

many analyses. It is also the bases of nonlinear analysis in most

of the cases.


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I I . Bar Element

Consider a uniform prismatic bar:

L length

A cross-sectional area

E elastic modulus

u u x= ( ) displacement

ε ε= ( ) x strain

σ σ= ( ) x stress

Strain-displacement relation:

ε =du

dx

(1)

Stress-strain relation:

σ ε= E (2)

i i j

ui u j

,E


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Stif fness Matr ix --- D irect Method

Assuming that the displacement u is varying linearly along

the axis of the bar, i.e.,

u x x

Lu

x

Lu

i j( ) = −

+1 (3)

we have

ε = −

=u u

L L

j i ∆(∆ = elongation) (4)

σ ε= = E E L

∆(5)

We also have

σ = F A

( F = force in bar) (6)

Thus, (5) and (6) lead to

F EA

Lk = =∆ ∆ (7)

where k EA

L= is the stiffness of the bar.

The bar is acting like a spring in this case and we concludethat element stiffness matrix is


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k = −

−

=

−

−

k k

k k

EA

L

EA

L EA

L

EA

L

or

k = −−

EA

L

1 1

1 1(8)

This can be verified by considering the equilibrium of the forces

at the two nodes. Element equilibrium equation is

EA

L

u

u

f

f

i

j

i

j

1 1

1 1

−−

=

(9)

Degree of Freedom (dof)

Number of components of the displacement vector at a

node.

For 1-D bar element: one dof at each node.

Physical Meaning of the Coefficients in k

The jth column of k (here j = 1 or 2) represents the forces

applied to the bar to maintain a deformed shape with unit

displacement at node j and zero displacement at the other node.


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Stif fness Matrix --- A F ormal Approach

We derive the same stiffness matrix for the bar using a

formal approach which can be applied to many other more

complicated situations.

Define two linear shape functions as follows

N N i j( ) , ( )ξ ξ ξ ξ= − =1 (10)

where

ξ ξ= ≤ ≤ x

L , 0 1 (11)

From (3) we can write the displacement as

u x u N u N ui i j j( ) ( ) ( ) ( )= = +ξ ξ ξ

or

[ ]u N N uui j

i

j

=

= Nu (12)

Strain is given by (1) and (12) as

ε = =

=du

dx

d

dxN u Bu (13)

where B is the element strain-displacement matrix, which is

[ ] [ ]B = = •d

dx N N

d

d N N

d

dxi j i j( ) ( ) ( ) ( )ξ ξ

ξ ξ ξ

ξ

i.e., [ ]B = −1 1/ / L L (14)


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Stress can be written as

σ ε= = E E Bu (15)

Consider the strain energy stored in the bar

( )

( )

U dV E dV

E dV

V V

V

= =

=

∫ ∫

∫

1

2

1

2

1

2

σ εT T T

T T

u B Bu

u B B u

(16)

where (13) and (15) have been used.

The work done by the two nodal forces is

W f u f ui i j j

= + =12

1

2

1

2u f

T (17)

For conservative system, we state that

U W = (18)

which gives

( )1

2

1

2u B B u u f

T T T E dV

V

∫

=

We can conclude that

( )B B u f T E dV V

∫

=

or


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ku f = (19)

where

( )k B BT= ∫ E dV V (20)

is the element stiffness matrix.

Expression (20) is a general result which can be used for

the construction of other types of elements. This expression can

also be derived using other more rigorous approaches, such as

the Principle of Minimum Potential Energy, or the Galerkin’s Method .

Now, we evaluate (20) for the bar element by using (14)

[ ]k = −

− = −−

∫

1

11 1

1 1

1 10

/

// /

L

L E L L Adx

EA

L

L

which is the same as we derived using the direct method.

Note that from (16) and (20), the strain energy in the

element can be written as

U =1

2u ku

T (21)


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Example 2.1

Problem: Find the stresses in the two bar assembly which is

loaded with force P , and constrained at the two ends,

as shown in the figure.

Solution: Use two 1-D bar elements.

Element 1,

u u

EA

L

1 2

1

2 1 1

1 1k =

−−

Element 2,

u u

EA

L

2 3

2

1 1

1 1k =

−−

Imagine a frictionless pin at node 2, which connects the twoelements. We can assemble the global FE equation as follows,

EA

L

u

u

u

F

F

F

2 2 0

2 3 1

0 1 1

1

2

3

1

2

3

−− −

−

=

1

2 A,E

2 3

,E 1 2


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Load and boundary conditions (BC) are,

u u F P 1 3 2

0= = =,

FE equation becomes,

EA

Lu

F

P

F

2 2 0

2 3 1

0 1 1

0

0

2

1

3

−− −

−

=

Deleting the 1st row and column, and the 3rd row and column,

we obtain,

[ ]{ } { } EA

Lu P 3

2 =

Thus,

u PL

EA2

3=

and

u

u

u

PL

EA

1

2

3

3

0

1

0

=

Stress in element 1 is

[ ]σ ε1 1 1 11

2

2 1

1 1

30

3

= = = −

= −

= −

=

E E E L Lu

u

E u u

L

E

L

PL

EA

P

A

B u / /


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Similarly, stress in element 2 is

[ ]σ ε2 2 2 22

3

3 2

1 1

03 3

= = = −

=

−= −

= −

E E E L Lu

u

E u u

L

E

L

PL

EA

P

A

B u / /

which indicates that bar 2 is in compression.

Check the results!

Notes:

• In this case, the calculated stresses in elements 1 and 2are exact within the linear theory for 1-D bar structures.

It will not help if we further divide element 1 or 2 into

smaller finite elements.

• For tapered bars, averaged values of the cross-sectionalareas should be used for the elements.

• We need to find the displacements first in order to findthe stresses, since we are using the displacement based

FEM .


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Example 2.2

Problem: Determine the support reaction forces at the two ends

of the bar shown above, given the following,

P E

A L =

= × = ×

= =

6 0 10 2 0 10

250 150

4 4

2

. , . ,

,

N N / mm

mm mm, 1.2 mm

2

∆

Solution:

We first check to see if or not the contact of the bar with

the wall on the right will occur. To do this, we imagine the wallon the right is removed and calculate the displacement at the

right end,

∆ ∆0

4

4

6 0 10 150

2 0 10 25018 12= = ×

× = > = PL

EA

( . )( )

( . )( ). .mm mm

Thus, contact occurs.

The global FE equation is found to be,

EA

L

u

u

u

F

F

F

1 1 0

1 2 1

0 1 1

1

2

3

1

2

3

−− −

−

=

1

,E

2 3

1 2

∆


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The load and boundary conditions are,

F P

u u

2

4

1 3

6 0 10

0 12

= = ×

= = =

.

, .

N

mm∆FE equation becomes,

EA

Lu

F

P

F

1 1 0

1 2 1

0 1 1

0

2

1

3

−− −

−

=

∆

The 2nd equation gives,

[ ] { } EA

L

u P 2 1

2−

=∆

that is,

[ ]{ } EA

Lu P

EA

L2

2 = +

∆

Solving this, we obtain

u PL

EA2

1

215= +

=∆ . mm

and

u

u

u

1

2

3

0

15

12

=

.

.

( )mm


37/169



To calculate the support reaction forces, we apply the 1st

and 3rd equations in the global FE equation.

The 1st equation gives,

[ ] ( ) F EA

L

u

u

u

EA

Lu

1

1

2

3

2

41 1 0 50 10= −

= − = − ×. N

and the 3rd equation gives,

[ ] ( ) F EA

L

uu

u

EA

Lu u

3

1

2

3

2 3

4

0 1 1

10 10

= −

= − +

= − ×. N

Check the results.!


38/169



Distr ibuted Load

Uniformly distributed axial load q (N/mm, N/m, lb/in) can

be converted to two equivalent nodal forces of magnitude qL/ 2.

We verify this by considering the work done by the load q,

[ ]

[ ]

[ ]

W uqdx u q Ld qL

u d

qL N N

u

ud

qLd

u

u

qL qL u

u

u uqL

qL

q

L

i j

i

j

i

j

i

j

i j

= = =

=

= −

=

=

∫ ∫ ∫ ∫

∫

1

2

1

2 2

2

21

1

2 2 2

1

2

2

2

0 0

1

0

1

0

1

0

1

( ) ( ) ( )

( ) ( )

/

/

ξ ξ ξ ξ

ξ ξ ξ

ξ ξ ξ

i

q

qL/2

i

qL/2


39/169



that is,

W qL

qLq

T

q q= =

1

2

2

2u f f with

/

/(22)

Thus, from the U=W concept for the element, we have

1

2

1

2

1

2u ku u f u f

T T T

q= + (23)

which yields

ku f f = +q (24)

The new nodal force vector is

f f + = +

+

q

i

j

f qL

f qL

/

/

2

2(25)

In an assembly of bars,

1 3

q

qL/2

1 3

qL/2

2

2

qL


40/169



Bar Elements in 2-D and 3-D Space

2-D Case

Local Global

x, y X, Y

u vi i' ', u vi i,

1 dof at node 2 dof’s at node

Note: Lateral displacement vi’ does not contribute to the stretch

of the bar, within the linear theory.

Transformation

[ ]

[ ]

u u v l m

u

v

v u v m l u

v

i i i

i

i

i i i

i

i

'

'

cos sin

sin cos

= + =

= − + = −

θ θ

θ θ

where l m= =cos , sinθ θ .

i

ui’ Y

θ

uivi


41/169



In matrix form,

u

v

l m

m l

u

v

i

i

i

i

'

'

=−

(26)

or,

u Tui i' ~=

where the transformation matrix

~T

= −

l m

m l (27)

is orthogonal , that is,~ ~T T

− =1 T .

For the two nodes of the bar element, we have

u

v

u

v

l m

m l

l m

m l

u

v

u

v

i

i

j

j

i

i

j

j

'

'

'

'

= −

−

0 0

0 0

0 0

0 0

(28)

or,

u Tu' = with T

T 0

0 T=

~

~ (29)

The nodal forces are transformed in the same way,

f Tf ' = (30)


42/169



Sti f fness Matri x in the 2-D Space

In the local coordinate system, we have

EA L

uu

f f

i

j

i

j

1 11 1

−−

=

'

'

'

'

Augmenting this equation, we write

EA

L

u

v

u

v

f

f

i

i

j

j

i

j

1 0 1 0

0 0 0 0

1 0 1 0

0 0 0 0

0

0

−

−

=

'

'

'

'

'

'

or,

k u f ' ' '=

Using transformations given in (29) and (30), we obtain

k Tu Tf ' =

Multiplying both sides by TT and noticing that TT T = I, we

obtain

T k Tu f T ' = (31)

Thus, the element stiffness matrix k in the global coordinate

system is

k T k T= T ' (32)

which is a 4×4 symmetric matrix.


43/169



Explicit form,

u v u v

EA

L

l lm l lmlm m lm m

l lm l lm

lm m lm m

i i j j

k =− −− −

− −

− −

2 2

2 2

2 2

2 2

(33)

Calculation of the directional cosines l and m:

l X X L

m Y Y L

j i j i= =

−= =

−cos , sinθ θ (34)

The structure stiffness matrix is assembled by using the element

stiffness matrices in the usual way as in the 1-D case.

Element Stress

σ ε= =

= −

E E u

u E

L L

l m

l m

u

v

u

v

i

j

i

i

j

j

B

'

'

1 1 0 0

0 0

That is,

[ ]σ = − −

E

Ll m l m

u

v

u

v

i

i

j

j

(35)


44/169



Example 2.3

A simple plane truss is made

of two identical bars (with E, A, and

L), and loaded as shown in thefigure. Find

1) displacement of node 2;

2) stress in each bar.

Solution:

This simple structure is usedhere to demonstrate the assembly

and solution process using the bar element in 2-D space.

In local coordinate systems, we have

k k 1 2

1 1

1 1

' '= −

−

= EA

L

These two matrices cannot be assembled together, because they

are in different coordinate systems. We need to convert them to

global coordinate system OXY.

Element 1:

θ

= = =45

2

2

o

l m,

Using formula (32) or (33), we obtain the stiffness matrix in the

global system

Y 1

2

45o

45o

3

2

1

1

2


45/169



u v u v

EA

L

T

1 1 2 2

1 1 1 12

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

k T k T= =

− −

− −

− −

− −

'

Element 2:

θ = = − =1352

2

2

2

o l m, ,

We have,

u v u v

EA

L

T

2 2 3 3

2 2 2 22

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

k T k T= =

− −

− −

− −

− −

'

Assemble the structure FE equation,

u v u v u v

EA

L

u

v

u

v

u

v

F

F

F

F

F

F

X

Y

X

Y

X

Y

1 1 2 2 3 3

1

1

2

2

3

3

1

1

2

2

3

3

2

1 1 1 1 0 0

1 1 1 1 0 0

1 1 2 0 1 1

1 1 0 2 1 1

0 0 1 1 1 1

0 0 1 1 1 1

− −

− −

− − −− − −

− −

− −

=


46/169



Load and boundary conditions (BC):

u v u v F P F P X Y 1 1 3 3 2 1 2 2

0= = = = = =, ,

Condensed FE equation,

EA

L

u

v

P

P 2

2 0

0 2

2

2

1

2

=

Solving this, we obtain the displacement of node 2,

u

v

L

EA

P

P

2

2

1

2

=

Using formula (35), we calculate the stresses in the two bars,

[ ] ( )σ11

2

1 2

2

21 1 1 1

0

0 2

2= − −

= + E

L

L

EA P

P

A P P

[ ] ( )σ2

1

2

1 2

2

21 1 1 1

0

0

2

2= − −

= − E

L

L

EA

P

P

A P P

Check the results :

Look for the equilibrium conditions, symmetry,

antisymmetry, etc.


47/169



Example 2.4 ( Multipoint Constraint )

For the plane truss shown above,

P L m E GPa

A m

A m

= = =

= ×

= ×

−

−

1000 1 210

6 0 10

6 2 10

4 2

4 2

kN,

for elements 1 and 2,

for element 3.

, ,

.

Determine the displacements and reaction forces.

Solution:

We have an inclined roller at node 3, which needs special

attention in the FE solution. We first assemble the global FE

equation for the truss.

Element 1:

θ = = =90 0 1o l m, ,

Y

45o

3

2

1

3

2

1

’

’


48/169



u v u v1 1 2 2

1

9 4210 10 6 0 10

1

0 0 0 0

0 1 0 1

0 0 0 0

0 1 0 1

k =

× × −

−

−( )( . )( ) N / m

Element 2:

θ = = =0 1 0o l m, ,

u v u v2 2 3 3

2

9 4210 10 6 0 10

1

1 0 1 0

0 0 0 0

1 0 1 0

0 0 0 0

k = × ×

−

−

−( )( . )( ) N / m

Element 3:

θ = = =451

2

1

2

o l m, ,

u v u v1 1 3 3

3

9 4210 10 6 2 10

2

05 0 5 05 05

05 0 5 05 05

0 5 05 0 5 0 5

0 5 05 0 5 0 5

k = × ×

− −

− −

− −− −

−( )( )

. . . .

. . . .

. . . .

. . . .

( ) N / m


49/169



The global FE equation is,

1260 10

0 5 05 0 0 05 05

15 0 1 05 05

1 0 1 0

1 0 0

15 05

05

5

1

1

2

2

3

3

1

1

2

2

3

3

×

− −

− − −

−

=

. . . .

. . .

. .

.Sym.

u

v

u

v

u

v

F

F

F

F

F

F

X

Y

X

Y

X

Y

Load and boundary conditions (BC):

u v v v

F P F X x

1 1 2 3

2 3

0 0

0

= = = =

= =

, ,

, .

'

'

and

From the transformation relation and the BC, we have

vu

vu v

3

3

3

3 3

2

2

2

2

2

20' ( ) ,= −

= − + =

that is,

u v3 3

0− =

This is a multipoint constraint (MPC).

Similarly, we have a relation for the force at node 3,

F F

F F F

x

X

Y

X Y 3

3

3

3 3

2

2

2

2

2

20

'( ) ,=

= + =

that is,

F F X Y 3 3

0+ =


50/169



Applying the load and BC’s in the structure FE equation by

‘deleting’ 1st, 2nd and 4th rows and columns, we have

1260 10

1 1 0

1 15 05

0 05 05

5

2

3

3

3

3

×

−

−

=

. .

. .

u

u

v

P

F

F

X

Y

Further, from the MPC and the force relation at node 3, the

equation becomes,

1260 10

1 1 0

1 15 05

0 05 05

5

2

3

3

3

3

×

−

−

=−

. .

. .

u

u

u

P

F

F

X

X

which is

1260 10

1 1

1 2

0 1

5 2

3

3

3

×

−

−

=

−

u

u

P

F

F

X

X

The 3rd equation yields,

F u X 3

5

31260 10= − ×

Substituting this into the 2nd equation and rearranging, we have

1260 101 1

1 3 05 2

3×

−

−

=

u

u

P

Solving this, we obtain the displacements,


51/169



u

u

P

P

2

3

5

1

2520 10

3 0 01191

0 003968

=×

=

.

.( )m

From the global FE equation, we can calculate the reactionforces,

F

F

F

F

F

u

u

v

X

Y

Y

X

Y

1

1

2

3

3

5

2

3

3

1260 10

0 0 5 05

0 0 5 05

0 0 0

1 15 05

0 05 05

500

500

0 0

500

500

= ×

− −

− −

−

=

−

−

−

. .

. .

. .

. .

. ( )kN

Check the results!

A general multipoint constraint (MPC) can be described as,

A u j j

j

=∑ 0

where A j’s are constants and u j’s are nodal displacement

components. In the FE software, such as MSC/NASTRAN ,

users only need to specify this relation to the software. The

software will take care of the solution.

Penalty Approach for Handli ng BC’ s and MPC’ s


52/169



3-D Case

Local Global

x, y, z X, Y, Z

u v wi i i' ' ', , u v wi i i, ,

1 dof at node 3 dof’s at node

Element stiffness matrices are calculated in the localcoordinate systems and then transformed into the global

coordinate system ( X, Y, Z ) where they are assembled.

FEA software packages will do this transformation

automatically.

Input data for bar elements:

• ( X, Y, Z ) for each node

• E and A for each element

i

Y

Z


53/169



I I I . Beam Element

Simple Plane Beam Element

L length

I moment of inertia of the cross-sectional area

E elastic modulus

v v x= ( ) deflection (lateral displacement) of theneutral axis

θ =dv

dx rotation about the z-axis

F F x= ( ) shear force

M M x= ( ) moment about z-axis

Elementary Beam Theory:

EI d v

dx M x

2

2 = ( ) (36)

σ = − My

I (37)

i

v j , F j

,I θi , M i θ j , M j

vi , F i


54/169



Di rect Method

Using the results from elementary beam theory to compute

each column of the stiffness matrix.

(Fig. 2.3-1. on Page 21 of Cook’s Book)

Element stiffness equation (local node: i, j or 1, 2):

v v

EI

L

L L

L L L L

L L

L L L L

v

v

F

M

F

M

i i j j

i

i

j

j

i

i

j

j

θ θ

θ

θ

3

2 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

−−

− − −

−

=

(38)


55/169



Formal Approach

Apply the formula,

k B B= ∫ T L

EI dx

0

(39)

To derive this, we introduce the shape functions,

N x x L x L

N x x x L x L

N x x L x L

N x x L x L

1

2 2 3 3

2

2 3 2

3

2 2 3 3

4

2 3 2

1 3 2

2

3 2

( ) / /

( ) / /

( ) / /

( ) / /

= − +

= − +

= −

= − +

(40)

Then, we can represent the deflection as,

[ ]

v x

N x N x N x N x

v

v

i

i

j

j

( )

( ) ( ) ( ) ( )

=

=

Nu

1 2 3 4

θ

θ

(41)

which is a cubic function. Notice that,

N N

N N L N x

1 3

2 3 4

1+ =+ + =

which implies that the rigid body motion is represented by the

assumed deformed shape of the beam.


56/169



Curvature of the beam is,

d v

dx

d

dx

2

2

2

2= =Nu Bu (42)

where the strain-displacement matrix B is given by,

[ ]B N= =

= − + − + − − +

d

dx N x N x N x N x

L

x

L L

x

L L

x

L L

x

L

2

2 1 2 3 4

2 3 2 2 3 2

6 12 4 6 6 12 2 6

" " " "( ) ( ) ( ) ( )

(43)

Strain energy stored in the beam element is

( ) ( )

U dV My

I E

My

I dAdx

M

EI

Mdxd v

dx

EI d v

dx

dx

EI dx

EI dx

T

V A

L T

T

L T L

T

L

T T

L

= = −

−

= =

=

=

∫ ∫ ∫

∫ ∫ ∫

∫

1

2

1

2

1

1

2

1 1

2

1

2

1

2

0

0

2

2

2

2

0

0

0

σ ε

Bu Bu

u B B u

We conclude that the stiffness matrix for the simple beam

element is

k B B= ∫ T L

EI dx

0


57/169



Applying the result in (43) and carrying out the integration, we

arrive at the same stiffness matrix as given in (38).

Combining the axial stiffness (bar element), we obtain the

stiffness matrix of a general 2-D beam element ,

u v u v

EA

L

EA

L EI

L

EI

L

EI

L

EI

L EI

L

EI

L

EI

L

EI

L EA

L

EA

L EI

L

EI

L

EI

L

EI

L

EI L

EI L

EI L

EI L

i i i j j jθ θ

k =

−

−

−

−

− − −

−

0 0 0 0

012 6

012 6

06 4

06 2

0 0 0 0

012 6

012 6

0 6 2 0 6 4

3 2 3 2

2 2

3 2 3 2

2 2

3-D Beam Element

The element stiffness matrix is formed in the local (2-D)

coordinate system first and then transformed into the global (3-D) coordinate system to be assembled.

(Fig. 2.3-2. On Page 24)


58/169



Example 2.5

Given: The beam shown above is clamped at the two ends and

acted upon by the force P and moment M in the mid-span.

Find : The deflection and rotation at the center node and the

reaction forces and moments at the two ends.

Solution: Element stiffness matrices are,

v v

EI

L

L L

L L L L

L L

L L L L

1 1 2 2

1 3

2 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

θ θ

k =

−

−

− − −

−

v v

EI

L

L L

L L L L

L L

L L L L

2 2 3 3

2 3

2 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

θ θ

k =

−−

− − −

−

1 2 ,I

Y

3

1 2


59/169



Global FE equation is,

v v v

EI

L

L L L L L L

L L

L L L L L

L L

L L L L

v

v

v

F M

F

M

F

M

Y

Y

Y

1 1 2 2 3 3

3

2 2

2 2 2

2 2

1

1

2

2

3

3

1

1

2

2

3

3

12 6 12 6 0 06 4 6 2 0 0

12 6 24 0 12 6

6 2 0 8 6 2

0 0 12 6 12 6

0 0 6 2 6 4

θ θ θ

θ

θ

θ

−

−

− − −

−

− − −

−

=

Loads and constraints (BC’s) are,

F P M M

v v

Y 2 2

1 3 1 30

= − =

= = = =

, ,

θ θ

Reduced FE equation,

EI

L L

v P

M 3 22

2

24 0

0 8

=

−

θ

Solving this we obtain,

v L

EI

PL

M

2

2

2

24 3θ

= −

From global FE equation, we obtain the reaction forces andmoments,


60/169



F

M

F

M

EI

L

L

L L

L

L L

v

P M L

PL M

P M L

PL M

Y

Y

1

1

3

3

3

2

2

2

2

12 6

6 2

12 6

6 2

1

4

2 3

2 3

=

−

−

− −

=

+

+

−

− +

θ

/

/

Stresses in the beam at the two ends can be calculated using the

formula,

σ σ= = − x My

I

Note that the FE solution is exact according to the simple beamtheory, since no distributed load is present between the nodes.

Recall that,

EI d v

dx M x

2

2 = ( )

and

dM

dxV V

dV

dxq q

=

=

(

(

- shear force in the beam)

- distributed load on the beam)

Thus,

EI d v

dxq x

4

4 = ( )

If q( x)=0, then exact solution for the deflection v is a cubic

function of x, which is what described by our shape functions.


61/169



Equivalent Nodal L oads of Distr ibuted Transverse Load

This can be verified by considering the work done by the

distributed load q.

i

q

qL/2

i

qL/2

qL2 /12qL2 /12

q

qL qL/2

qL2 /12


62/169



Example 2.6

Given: A cantilever beam with distributed lateral load p as

shown above.

Find : The deflection and rotation at the right end, the

reaction force and moment at the left end.

Solution: The work-equivalent nodal loads are shown below,

where

f pL m pL= =/ , /2 122

Applying the FE equation, we have

1 2 ,I

1 2 ,I

m


63/169



EI

L

L L

L L L L

L L

L L L L

v

v

F

M

F

M

Y

Y

3

2 2

2 2

1

1

2

2

1

1

2

2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

−

−

− − −

−

=

θ

θ

Load and constraints (BC’s) are,

F f M m

v

Y 2 2

1 10

= − =

= =

,

θ

Reduced equation is,

EI

L

L

L L

v f

m3 22

2

12 6

6 4

−

−

= −

θ

Solving this, we obtain,

v L

EI

L f Lm

Lf m

pL EI

pL EI

2

2

2 4

3

6

2 3

3 6

8

6θ

= − +

− +

= −

−

/

/

(A)

These nodal values are the same as the exact solution.

Note that the deflection v( x) (for 0


64/169



The errors in (B) will decrease if more elements are used. The

equivalent moment m is often ignored in the FEM applications.

The FE solutions still converge as more elements are applied.

From the FE equation, we can calculate the reaction forceand moment as,

F

M

L

EI

L

L L

v pL

pL

Y 1

1

3

2

2

2

2

12 6

6 2

2

5 12

= −

−

=

θ

/

/

where the result in (A) is used. This force vector gives the total

effective nodal forces which include the equivalent nodal forcesfor the distributed lateral load p given by,

−

−

pL

pL

/

/

2

122

The correct reaction forces can be obtained as follows,

F

M

pL

pL

pL

pL

pL

pLY 1

1

2 2 2

2

5 12

2

12 2

=

−

−

−

=

/

/

/

/ /

Check the results!


65/169



Example 2.7

Given: P = 50 kN, k = 200 kN/m, L = 3 m,

E = 210 GPa, I = 2×10-4 m4.

Find : Deflections, rotations and reaction forces.

Solution:

The beam has a roller (or hinge) support at node 2 and a

spring support at node 3. We use two beam elements and one

spring element to solve this problem.

The spring stiffness matrix is given by,

v v

k k

k k s

3 4

k = −

−

Adding this stiffness matrix to the global FE equation (see

Example 2.5), we have

12

,I

Y

3

1 2

k

4


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v v v v

EI

L

L L

L L L

L

L L L

k L

L

k

Symmetry k

v

v

v

v

F

M

F

M

F

M

F

Y

Y

Y

Y

1 1 2 2 3 3 4

3

2 2

2 2

2

1

1

2

2

3

3

4

1

1

2

2

3

3

4

12 6 12 6 0 0

4 6 2 0 0

24 0 12 6

8 6 2

12 6

4

0

0

0

0

0

θ θ θ

θ

θ

θ

−

−

−

−

+ − −

=

' '

'

in which

k L

EI k '=

3

is used to simply the notation.

We now apply the boundary conditions,

v v v M M F P

Y

1 1 2 4

2 3 3

00

= = = == = = −θ ,

,

‘Deleting’ the first three and seventh equations (rows and

columns), we have the following reduced equation,

EI

L

L L L

L k L L L L

v P 3

2 2

2 2

2

3

3

8 6 2

6 12 62 6 4

0

0

−

− + −

−

= −

'

θ

θ

Solving this equation, we obtain the deflection and rotations at

node 2 and node 3,


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θ

θ

2

3

3

2

12 7

3

7

9

v PL

EI k L

= −+

( ' )

The influence of the spring k is easily seen from this result.

Plugging in the given numbers, we can calculate

θ

θ

2

3

3

0 002492

0 01744

0 007475

v

=

−

−

−

.

.

.

rad

m

rad

From the global FE equation, we obtain the nodal reaction

forces as,

F

M

F

F

Y

Y

Y

1

1

2

4

69 78

69 78

1162

3488

=

−

− ⋅

.

.

.

.

kN

kN m

kN

kN

Checking the results : Draw free body diagram of the beam

1 2

50 kN

3

3.488 kN116.2 kN

69.78 kN

69.78 kN⋅m


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Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems


Chapter 3. Two-Dimensional Problems

I . Review of the Basic Theory

In general, the stresses and strains in a structure consist of

six components:

σ σ σ τ τ τ x y z xy yz zx, , , , , for stresses,

and

ε ε ε γ γ γ x y z xy yz zx, , , , , for strains.

Under contain conditions, the state of stresses and strains

can be simplified. A general 3-D structure analysis can,

therefore, be reduced to a 2-D analysis.

z

x

σ y

σ z

yz

zx

xy


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Plane (2-D) Problems

• Plane stress:

σ τ τ ε z yz zx z = = = ≠0 0( ) (1)

A thin planar structure with constant thickness and

loading within the plane of the structure ( xy-plane).

• Plane strain:ε γ γ σ z yz zx z = = = ≠0 0( ) (2)

A long structure with a uniform cross section and

transverse loading along its length ( z-direction).

z

z


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Stress-Strain-Temperature (Consti tutive) Relations

For elastic and isotropic materials, we have,

εε

γ

ν ν

σσ

τ

εε

γ

x

y

xy

x

y

xy

x

y

xy

E E E E

G

=−

−

+

1 01 0

0 0 1

0

0

0

/ // /

/

(3)

or,

ε σ ε= +−E 1 0

where ε0 is the initial strain, E the Young’s modulus, ν the

Poisson’s ratio and G the shear modulus. Note that,

G E

=+2 1( ) ν

(4)

which means that there are only two independent materials

constants for homogeneous and isotropic materials.We can also express stresses in terms of strains by solving

the above equation,

σ

σ

τ ν

ν

ν

ν

ε

ε

γ

ε

ε

γ

x

y

xy

x

y

xy

x

y

xy

E

=

−−

−

1

1 0

1 0

0 0 1 2

2

0

0

0( ) /

(5)

or,

σ ε σ= +E 0

where σ ε0 0= −E is the initial stress.


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The above relations are valid for plane stress case. For

plane strain case, we need to replace the material constants in

the above equations in the following fashion,

E E

G G

→−

→−

→

1

1

2 ν

ν ν

ν(6)

For example, the stress is related to strain by

σ

σ

τ ν ν

ν ν

ν ν

ν

ε

ε

γ

ε

ε

γ

x

y

xy

x

y

xy

x

y

xy

E

=+ −

−−

−

−

( )( ) ( ) /

1 1 2

1 0

1 0

0 0 1 2 2

0

0

0

in the plane strain case.

Initial strains due to temperature change (thermal loading)

is given by,

ε

ε

γ

α

α

x

y

xy

T

T

0

0

00

=

∆∆ (7)

where α is the coefficient of thermal expansion, ∆T the changeof temperature. Note that if the structure is free to deform under

thermal loading, there will be no (elastic) stresses in the

structure.


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Strain and Displacement Relations

For small strains and small rotations, we have,

ε ∂∂

ε ∂∂

γ ∂∂

∂∂

x y xyu x

v y

u y

v x

= = = +, ,

In matrix form,

ε

ε

γ

∂ ∂

∂ ∂

∂ ∂ ∂ ∂

x

y

xy

x

y

y x

u

v

=

/

/

/ /

0

0 , or ε = Du (8)

From this relation, we know that the strains (and thus

stresses) are one order lower than the displacements, if the

displacements are represented by polynomials.

Equil ibrium Equations

In elasticity theory, the stresses in the structure must satisfy

the following equilibrium equations,

∂σ

∂

∂τ

∂

∂τ

∂

∂σ

∂

x xy

x

xy y

y

x y f

x y f

+ + =

+ + =

0

0

(9)

where f x and f y are body forces (such as gravity forces) per unit

volume. In FEM, these equilibrium conditions are satisfied in

an approximate sense.


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Boundary Conditions

The boundary S of the body can be divided into two parts,

S u and S t . The boundary conditions (BC’s) are described as,

u u v v S

t t t t S

u

x x y y t

= == =

, ,

, ,

on

on(10)

in which t x and t y are traction forces (stresses on the boundary)

and the barred quantities are those with known values.

In FEM, all types of loads (distributed surface loads, bodyforces, concentrated forces and moments, etc.) are converted to

point forces acting at the nodes.

Exact Elastici ty Solution

The exact solution (displacements, strains and stresses) of a

given problem must satisfy the equilibrium equations (9), the

given boundary conditions (10) and compatibility conditions

(structures should deform in a continuous manner, no cracks and

overlaps in the obtained displacement fields).

t x

t y

S u

S t


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Example 3.1

A plate is supported and loaded with distributed force p as

shown in the figure. The material constants are E and ν.

The exact solution for this simple problem can be found

easily as follows,

Displacement:

u p

E x v p

E y= = −, ν

Strain:

ε ε ν γ x y xy

p

E

p

E = = − =, , 0

Stress:σ σ τ x y xy p= = =, ,0 0

Exact (or analytical) solutions for simple problems are

numbered (suppose there is a hole in the plate!). That is why we

need FEM!


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I I . F ini te Elements for 2-D Problems

A General Formula for the Stiff ness Matrix

Displacements (u, v) in a plane element are interpolated

from nodal displacements (ui, vi) using shape functions N i as

follows,

uv

N N N N

u

v

u

v

=

=1 2

1 2

1

1

2

2

0 00 0

L

L

M

or u Nd (11)

where N is the shape function matrix, u the displacement vector

and d the nodal displacement vector. Here we have assumed

that u depends on the nodal values of u only, and v on nodalvalues of v only.

From strain-displacement relation (Eq.(8)), the strain vector

is,

ε ε= = =Du DNd Bd, or (12)

where B = DN is the strain-displacement matrix.


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Consider the strain energy stored in an element,

( )

( )

U dV dV

dV dV

dV

T

V

x x y y xy xy

V

T

V

T

V

T T

V

T

= = + +

= =

=

=

∫ ∫

∫ ∫

∫

1

2

1

2

1

2

1

2

1

2

1

2

σ ε σ ε σ ε τ γ

ε ε ε εE E

d B EB d

d kd

From this, we obtain the general formula for the element

stiffness matrix,

k B EB= ∫ T V

dV (13)

Note that unlike the 1-D cases, E here is a matrix which is given by the stress-strain relation (e.g., Eq.(5) for

Finite Element Methods Lectures_Lui.pdf

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